L3 2 1Mesh Analysis

Introduces Mesh Current method of analyzing Circuits

In this next set of videos,

we're going to introduce a circuit
analysis method known as mesh analysis.

My name is Lee Brinton,

I'm an Electrical Engineering Instructor
at Salt Lake Community College.

In this videos we'll make
the distinction between a loop and

a mesh, we'll make distinctions between
branch currents and mesh currents

We'll demonstrate the process of
analyzing circuits using mesh analysis.

Look at how dependent sources
impact our analysis approach.

And we'll also introduce
concept of supermeshes.

We're gonna find that a supermesh is
analagous to a supernode that we ran into

doing node analysis First of all
the difference between a loop and a mesh.

A loop is any closed path.

In this circuit we have a closed
path here, a closed path there,

and yet one more closed path.

So this circuit has three different loops.

Let us formally define a mesh now.

A mesh is a closed Path that contains
no other closed paths within it.

So in this case, we have a mesh here,

a mesh here, but this outer loop,

because it contains two loops within it,
does not qualify as a mesh.

Lets also make the distinction between
branch currents and mesh currents.

As we know,
a branch current is the current flowing in

any single branch of a circuit.

This circuit has three different branches,
there is this branch here,

I sub a, this branch here that has
the current to find I sub b, and

this branch here that
has the current I sub c.

A mesh current is somewhat different.

A mesh current is something of a contrived
or a tool that we had define to

assist in this analysis and
to understand it think simply about or

imagine that there was a current
That flowed around this loop.

That is a mesh current.

Similarly imagine that there’s
a current flowing around this loop.

Those two mesh currents.

We're now gonna see that those two miss
current are all that we need to represent

the minimum number of variables that
we need to analyze the circuit.

So what the difference between i1 and i2?

Don't they both impact this branch?

And in fact they do.

Is defined as the current going down r3,
it's a branch current.

In terms of our mesh currents, I sub b

would equal our mesh current I1 flowing
down in the same direction as I sub b

minus the I2 mesh current that's
flowing or opposing that.

So I sub b then would equal I1 minus I2.

On the other hand,

I1 is this current flowing through here.

And that current is in fact
the branch current Ia.

So we can say, then, that Ia = I1.

Or I guess more appropriately,
more intuitively,

we'd say that I1 is simply
the branch current Ia.

And similarly over here,
this mesh current I2 Is

the same as the branch current I sub c.

So we can then see that I
sub 2 is equal to I sub c.

Thus, we can define any branch
current in this circuit in terms of

the mesh currents I1 and I2.

Now let's see how we

would go about writing
The equations to solve this,

or to analyze this kind of circuit.In
node analysis we write KCL equations.

One KCL for each node.

In mesh analysis,
we're going to write mesh equations or

KVL equations, summing the voltage
drops around the closed loops.

One for each mesh.

So for example,
let's start with this left hand mesh, and

write a KVL equation starting right here.

a we go across the voltage source,
we're increasing V0 volt, and we're

gonna sum the voltage drops, therefore
the voltage increase would be negative.

And we have then -Vo continuing on around
here the full stop across that resistor.

In terms of I1 and I2,
our mesh currency is going to be,

and because we're going
in the direction of I1,

it'll be a plus
R1I1 Continuing on down here,

the voltage drop across R3 in
terms of mesh currency is going to

be R3 times the current flowing in
that direction, which is I1 minus I2.

And the sum of those three
terms must equal zero.

Now look at the right-hand mesh.

Starting right here, and let's go in
the direction that I2 has referenced.

We have, then,
the voltage drop going plus to minus,

going in the voltage, direction of
current flow, we will have R3 times

the current going up that
branch Which is (I2- I1).

Continuing on around this
loop we have the voltage

drop across R2 is going
to be + R2 times I2 and

the sum of those two
terms then must equal 0

We have two equations with two unknowns,
we've got everything we need to solve so

the next step with then simply
be to combine like terms.

The first equation, lets factor or
combine the I1 terms and factor I1 out.

In doing so we're left with an R1 + R3.

Similarly, for I2,
there's only one I2 term.

It's got a -R3 times that.

So -R3 there.

Bringing this -V0 to the other
side of the positive,

we have on the left hand side = V0.

The second equation will
similarly factor out I1.

When we do, we got a negative R3

now working with I2 we have, what is that?

R2 plus R3.

And the sum of those terms has to equal 0.

So, we have two equations, two unknowns
if we knew what R1, R2, and R3 and

V0 are we could plug in those values and
we would be able t solve for I1.

And I2.

Given those two values, we could then go
back and calculate any branch voltage or

branch currents that we were
interested in in this circuit.